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人気のある三角関数 >

2cos^2(x)+sin(x)=2

解

2 cos^{2} (x) + sin (x) = 2

解

x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

解答ステップ

2 cos^{2} (x) + sin (x) = 2

両辺から

2

を引く

2 cos^{2} (x) + sin (x) - 2 = 0

三角関数の公式を使用して書き換える

- 2 + sin (x) + 2 cos^{2} (x)

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 2 + sin (x) + 2 (1 - sin^{2} (x))

簡素化

- 2 + sin (x) + 2 (1 - sin^{2} (x)) : sin (x) - 2 sin^{2} (x)

- 2 + sin (x) + 2 (1 - sin^{2} (x))

拡張

2 (1 - sin^{2} (x)) : 2 - 2 sin^{2} (x)

2 (1 - sin^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

= 2 \cdot 1 - 2 sin^{2} (x)

数を乗じる:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (x)

= - 2 + sin (x) + 2 - 2 sin^{2} (x)

簡素化

- 2 + sin (x) + 2 - 2 sin^{2} (x) : sin (x) - 2 sin^{2} (x)

- 2 + sin (x) + 2 - 2 sin^{2} (x)

条件のようなグループ

= sin (x) - 2 sin^{2} (x) - 2 + 2

- 2 + 2 = 0

= sin (x) - 2 sin^{2} (x)

= sin (x) - 2 sin^{2} (x)

= sin (x) - 2 sin^{2} (x)

sin (x) - 2 sin^{2} (x) = 0

置換で解く

sin (x) - 2 sin^{2} (x) = 0

仮定:

sin (x) = u

u - 2 u^{2} = 0

u - 2 u^{2} = 0 : u = 0, u = \frac{1}{2}

u - 2 u^{2} = 0

標準的な形式で書く

a x^{2} + b x + c = 0

- 2 u^{2} + u = 0

解くとthe二次式

- 2 u^{2} + u = 0

二次Equationの公式:

次の場合:

a = - 2, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 0 = 1

1^{2} - 4 (- 2) \cdot 0

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 0

規則を適用

- (- a) = a

= 1 + 4 \cdot 2 \cdot 0

規則を適用

0 \cdot a = 0

= 1 + 0

数を足す:

1 + 0 = 1

= 1

規則を適用

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- 1 + 1}{2 ( - 2 )}, u_{2} = \frac{- 1 - 1}{2 ( - 2 )}

u = \frac{- 1 + 1}{2 ( - 2 )} : 0

\frac{- 1 + 1}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 + 1}{- 2 \cdot 2}

数を足す/引く:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{0}{- 4}

分数の規則を適用する:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{4}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 1 - 1}{2 ( - 2 )} : \frac{1}{2}

\frac{- 1 - 1}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 - 1}{- 2 \cdot 2}

数を引く:

- 1 - 1 = - 2

= \frac{- 2}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

分数の規則を適用する:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

二次equationの解:

u = 0, u = \frac{1}{2}

代用を戻す

u = sin (x)

sin (x) = 0, sin (x) = \frac{1}{2}

sin (x) = 0, sin (x) = \frac{1}{2}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

以下の一般解

sin (x) = 0

sin (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

以下の一般解

sin (x) = \frac{1}{2}

sin (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

すべての解を組み合わせる

x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

グラフ

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